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对于任何的文件名的操作,你都应该使用 os.path 模块

Open limboinf opened this issue 8 years ago • 0 comments

eg:

>>> import os
>>> path = '/Users/beazley/Data/data.csv'

>>> # 获取路径的最后一个组件
>>> os.path.basename(path)
'data.csv'

>>> # 获取目录名称
>>> os.path.dirname(path)
'/Users/beazley/Data'

>>> # 组合路径
>>> os.path.join('tmp', 'data', os.path.basename(path))
'tmp/data/data.csv'

>>> # 扩展用户的主目录
>>> path = '~/Data/data.csv'
>>> os.path.expanduser(path)
'/Users/beazley/Data/data.csv'

>>> # Split the file extension
>>> os.path.splitext(path)
('~/Data/data', '.csv')

limboinf avatar Jan 13 '16 15:01 limboinf